A line is a common tangent to the circle (x - 3) ^ 2 + y ^ 2 = 9 and the parabola y ^ 2 = 4x . If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to ___ .

Correct answer: 0 Let coordinate of point A( t ^ 2 , 2t)(∵ a = 1) equation of tangent at point A yt = x + t ^ 2 x - ty + t ^ 2 = 0 centre of circle (3, 0) Now Perpendicular distance = radius | 3 - 0 + t ^ 2 1 + t ^ 2 | = 3 (3 + t ^ 2 ) ^ 2 = 9(1 + t ^ 2 ) 9 + t ^ 4 + 6 t ^ 2 = 9 + 9 t ^ 2 t = 0, - 3 , 3 So point A(3, 2 3 ) ⇒ a = 3, b = 2 3 (Since it lies in first quadrant) For point B which is foot of perpendicular from centre (3, 0) to the tangent x - 3 y + 3 = 0 (c - 3)/(1) = d - 0 - 3 = (- (3 - 0 + 3))/(4) ⇒ c = (3)/(2)d = 3 3 2 ⇒ 2(3 + (3)/(2)) = 9

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Parabola

A line is a common tangent to the circle (x - 3) ^ 2 + y ^ 2 = 9...

A line is a common tangent to the circle

{\left(x - 3\right)}^{2} + {y}^{2} = 9 and the parabola

{y}^{2} = 4x

. If the two points of contact \left(a, b\right) and \left(c, d\right)

are distinct and lie in the first quadrant, then

2\left(a + c\right)

is equal to ___ . A line is a common tangent to the circle (x - 3) ^ 2 + y ^ 2 = 9 and the parabola y ^ 2 = 4x . If the two points of

Correct answer

Step-by-step solution

Let coordinate of point $A\left({t}^{2}, 2t\right)\left(∵ a = 1\right)$

equation of tangent at point $A$

$yt = x + {t}^{2}$

$x - ty + {t}^{2} = 0$

A line is a common tangent to the circle (x - 3) ^ 2 + y ^ 2 = 9 and the parabola y ^ 2 = 4x . If the two points of

centre of circle $\left(3, 0\right)$

Now Perpendicular distance = radius

\left|\dfrac{3 - 0 + {t}^{2}}{\sqrt{1 + {t}^{2}}}\right| = 3 $$ {\left(3 + {t}^{2}\right)}^{2} = 9\left(1 + {t}^{2}\right) $$ 9 + {t}^{4} + 6{t}^{2} = 9 + 9{t}^{2} $$ t = 0, - \sqrt{3}, \sqrt{3} $So point$ A\left(3, 2\sqrt{3}\right) $$ ⇒ a = 3, b = 2\sqrt{3} $(Since it lies in first quadrant)For point$ B $which is foot of perpendicular from centre \left(3, 0\right)$ to the tangent $x - \sqrt{3}y + 3 = 0$

$\dfrac{c - 3}{1} = \dfrac{d - 0}{- \sqrt{3}} = \dfrac{- \left(3 - 0 + 3\right)}{4}$

$⇒ c = \dfrac{3}{2}d = \dfrac{3\sqrt{3}}{2}$

$⇒ 2\left(3 + \dfrac{3}{2}\right) = 9$

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